An acid ${\left({\mathrm{H}}_2\mathrm{A}\right)}_{\mathrm{l}}$reacts with base $(\mathrm{BOH}{)}_{\mathrm{l}}$ . Initially H₂A is mixed with BOH in molar ratio 1 : X (X > 1). The system reached equilibrium as shown below at 298K
$\begin{array}{l}{\mathrm{H}}_2\mathrm{A}+\mathrm{BOH}\rightleftharpoons \mathrm{BHA}+{\mathrm{H}}_2\mathrm{O}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{K}}_1=\frac{\left[\mathrm{BHA}\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{H}}_2\mathrm{A}\right]\left[\mathrm{BOH}\right]}=20\\ \mathrm{BHA}+\mathrm{BOH}\rightleftharpoons {\mathrm{B}}_2\mathrm{A}+{\mathrm{H}}_2\mathrm{O}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{K}}_2=\frac{\left[{\mathrm{B}}_2\mathrm{A}\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[\mathrm{BHA}\right]\left[\mathrm{BOH}\right]}=20\end{array}$
Note : ${\mathrm{H}}_2\mathrm{A},\mathrm{BOH},\mathrm{BHA}\text{ and }{\mathrm{H}}_2\mathrm{O}$ are in undissociated forms.